Chapter 11: Problem 93
$$ \cos (-A+B+C)+\cos (A-B+C)+\cos (A+B-C)+\cos (A+B+C)=4 \cos A \cos B \cos C $$
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none}
Learning Materials
Features
Discover
Chapter 11: Problem 93
$$ \cos (-A+B+C)+\cos (A-B+C)+\cos (A+B-C)+\cos (A+B+C)=4 \cos A \cos B \cos C $$
These are the key concepts you need to understand to accurately answer the question.
All the tools & learning materials you need for study success - in one app.
Get started for free
$$ \sin 50^{\circ}-\sin 70^{\circ}+\sin 10^{\circ}=0 $$
$$ \cos 36^{\circ} \cos 42^{\circ} \cos 78^{\circ}=\frac{1}{8} $$
$$ \sin \frac{\theta}{2} \sin \frac{7 \theta}{2}+\sin \frac{3 \theta}{2} \sin \frac{11 \theta}{2}=\sin 2 \theta \sin 5 \theta \text { . } $$
$$ \frac{\sin (A-B)}{\cos A \cos B}+\frac{\sin (B-C)}{\cos B \cos C}+\frac{\sin (C-A)}{\cos C \cos A}=0 $$
$$ \frac{\sin (4 A-2 B)+\sin (4 B-2 A)}{\cos (4 A-2 B)+\cos (4 B-2 A)}=\tan (A+B) $$
What do you think about this solution?
We value your feedback to improve our textbook solutions.